Given
x,y ∈ [0,2c]
f(x),g(y) ∈ [0,2c]
Show
| xy-f(x)-g(y)|≥ c²
exist
I don't really get this question.
Please help me
It is assumed that $\displaystyle c>0$:
Consider the points $\displaystyle (0,2c),(2c,0),(0,0)$ if these work then we are finished. Otherwise,
$\displaystyle |f(0)+g(0)|<c^2$
$\displaystyle |f(2c)+g(0)|<c^2$
$\displaystyle |f(0)+g(2c)|<c^2$
Now, the claim is that $\displaystyle |f(2c)+g(2c)-4c^2|\geq c^2$.
This is true because,
$\displaystyle f(2c)+g(2c) = [f(2c)+g(0)] + [f(0)+g(2c)] - [f(0)+g(0)] < c^2+c^2+c^2 = 3c^2$
But then,
$\displaystyle f(2c)+g(2c) - (2c)(2c) < 3c^2 - (2c)(2c) = -c^2$
Thus,
$\displaystyle |f(2c)+g(2c) - (2c)(2c)|\leq c^2$